What is the #"pH"# for an aqueous solution with a hydroxide concentration of #5 xx 10^(-3) "M"# at #25^@ "C"#?

2 Answers
Dec 25, 2017

Answer:

#11.7#

Explanation:

#"pOH" = - log["OH"^-]#

#"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30#

Now,

#"pH" + "pOH" = 14#

#"pH" = 14 - "pOH" = 14 - 2.30 = 11.7#

Answer:

#"pH" ~~ color(blue)(11.7)#

Explanation:

#color (indigo)(pH=-log [H_3O^+])#
#color (indigo)(pOH=-log [OH^-])#
#color(cyan)([OH^-]=0.005)#
Here,
#pOH=-log [0.005]#
#pOH=-log [5×10^-3]#
#pOH=2.3010299957#

#color (green)(pH=14-pOH)#

#ph=14-2.301 =11.699# #~~# #bb11.7#